If $$f(x)$$ is differentiable and $$\int\limits_0^{{t^2}} {xf\left( x \right)dx = {2 \over 5}{t^5},} $$ then $$f\left( {{4 \over {25}}} \right)$$ equals

If $$l\left( {m,n} \right) = \int\limits_0^1 {{t^m}{{\left( {1 + t} \right)}^n}dt,} $$ then the expression for $$l(m, n)$$ in terms of $$l(m+n, n-1)$$ is

If $$f\left( x \right) = \int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}} dt,$$ then $$f(x)$$ increases in

The area bounded by the curves $$y = \sqrt x ,2y + 3 = x$$ and
$$x$$-axis in the 1^st quadrant is